Proving $|\int_{0}^{x}{f(t)\,dt|^2} \le 2x^{1/2}\int_{0}^{x}{t^{1/2}| f(t)|^2\,dt}$ for $f \in L^2[0,\infty)$ and $x \geq 0$

Question

I'm currently in my real analysis class going over Hilbert spaces. My problem is proving that given $f \in L^2[0,\infty)$ and $x \geq 0$

$|\int_{0}^{x}{f(t)\,dt|^2} \le 2x^{1/2}\int_{0}^{x}{t^{1/2}| f(t)|^2\,dt}$

The problem states it establishes a special case of Hardy’s Inequalities. I know to use the Cauchy-Schwarz Inequality as a hint.

Edited. I forgot a square

Answer 1

This can be solved by a clever use of the Cuachy Schwarz Inequality by selecting our two functions cleverly. We state the inequality as

$$\left|\int_0^xg(t)h(t)dt\right|^2\leq\left(\int_0^xg(t)^2dt\right)\left(\int_0^xh(t)^2dt\right)$$

for square integrable functions $g$ and $h$. We will let $g=t^{1/4}f$ and $h=1/t^{1/4}$ (can you show that both of these functions are square integrable?). This tells us that

$$\left|\int_0^xf(t)dt\right|^2\leq\left(\int_0^xt^{1/2}|f(t)|^2dt\right)\left(\int_0^x\frac{1}{\sqrt{t}}dt\right)$$

And the rest is history.

Answer 2

$\int f(t)\, dt =\int t^{-1/4} (t^{1/4}f(t))\, dt$. Apply Holder's/Cauchy-Schwarz inequality now.